Category: mathematics

Mathematicians discover fractal-like probabilistic structures governing the distribution of prime numbers — a breakthrough 2,000 years in the making.

The distribution of prime numbers follows newly discovered probabilistic patterns, combining chaos theory and fractal geometry. (Scientific American, 2025)

Prime numbers are the atoms of arithmetic — the indivisible building blocks from which every whole number is constructed through multiplication. They begin simply enough: 2, 3, 5, 7, 11, 13… But their distribution along the number line has baffled mathematicians for millennia. They seem to appear randomly, with no obvious pattern, yet they are entirely determined by logic. How can something so orderly feel so chaotic?

This tension between determinism and apparent randomness has driven some of the deepest mathematics ever created, from Euler’s product formula in the 18th century to Riemann’s hypothesis in 1859 — a conjecture about the zeros of the Riemann zeta function that remains unproven and carries a $1 million prize for its resolution. In 2025, a new layer was peeled back: mathematicians discovered a set of probabilistic patterns governing how primes are distributed, patterns that involve both random chaotic behaviour and fractal geometry.

Fractals in the number line

The key discovery, highlighted by Scientific American in its year-end review of the top ten mathematical breakthroughs of 2025, is that the gaps and clustering of prime numbers — when viewed at large scales — follow statistical laws that resemble fractal structures. That is, the patterns are self-similar: zoom in or zoom out, and the same types of distributions appear. This is not the same as claiming primes are fractal (they are discrete and deterministic), but rather that their statistical fingerprint has fractal characteristics.

“Discovering new primes is difficult as you get to larger numbers. But in 2025, mathematicians found probabilistic patterns governing how primes are distributed — patterns involving fractals.” — Scientific American, 2025

A refined prime-counting method

Separately, a new and more precise technique for estimating the prime-counting function pi(x) — the number of primes up to a given value x — was developed in 2025. The method combines sieve-based elimination (filtering out composite numbers) with improved error corrections that reduce the gap between the estimate and the true count. While the method does not solve the Riemann Hypothesis, it narrows the uncertainty in prime counting to a degree that was previously out of reach, with practical applications in cryptography, where the distribution of large primes determines the security of encryption systems.

Sources & Further Reading

Scientific American (2025). The Top 10 Math Discoveries of 2025. scientificamerican.com

Quanta Magazine (2025). Year in Review: Mathematics. quantamagazine.org

Medium / Wahlastore15 (2025). Mathematics in 2025: Breakthroughs That Redefined the Field. medium.com

Entechonline (2026). Top 10 Mathematics Discoveries in 2025. entechonline.com

How mathematicians proved that a needle rotated through every direction in 3D space can do so in a set of zero volume.

A rotating needle sweeps through all directions — the Kakeya set question asks for the smallest possible “sweep area.” Settled in 3D by Wang & Zahl, 2025.

Imagine a needle — a thin line segment of length one — lying flat on a table. You want to rotate it through every possible direction in the plane, completing a full 360-degree turn. The region swept out by the needle must have some area. But how small can that area be? Intuitively, you might think it must be at least something reasonably large. Remarkably, in two dimensions, mathematicians proved in the 1920s that you can rotate a needle through every direction in a region of arbitrarily small area. These minimal regions are called Kakeya sets.

The Kakeya conjecture asks a harder question in higher dimensions: in n-dimensional space, what is the fractal dimension of a Kakeya set — a set that contains a unit line segment in every possible direction? In two dimensions, the answer is known: 2. The three-dimensional case, however, resisted proof for nearly a century. Until 2025.

Wang and Zahl settle 3D Kakeya

In early 2025, Hong Wang (New York University) and Joshua Zahl (University of British Columbia) published a proof of the three-dimensional Kakeya conjecture. Their result establishes that a Kakeya set in three-dimensional space must have Hausdorff dimension 3 — that is, it must be “full-dimensional” even if it has zero volume. The proof runs to over 100 pages and draws on sophisticated techniques from harmonic analysis and additive combinatorics.

“The three-dimensional Kakeya problem sat unsolved for nearly a century, and its resolution is one of the landmark achievements of 2025.” — Quanta Magazine

Beyond the abstract

The Kakeya problem might sound purely theoretical, but it has deep connections to some of the most important unsolved problems in mathematics, including the Riemann Hypothesis and the behaviour of the Fourier transform in higher dimensions. Progress on Kakeya-type problems has historically driven breakthroughs in signal processing and the mathematics of waves — areas that underpin everything from medical imaging to wireless communication. Wang and Zahl’s result is not merely an isolated geometric curiosity; it opens new avenues of attack on problems that have blocked progress in analysis for decades.

Sources & Further Reading

Wang, H. & Zahl, J. (2025). Three-dimensional Kakeya conjecture: proof. Annals of Mathematics (preprint).

Quanta Magazine (2025). The Biggest Breakthroughs in Mathematics: 2025. quantamagazine.org

Scientific American (2025). The Top 10 Math Discoveries of 2025. scientificamerican.com

Medium (2025). Mathematics in 2025: Breakthroughs That Redefined the Field. medium.com

A 125-year-old challenge by David Hilbert — to mathematically unify the laws of physics — just got a major breakthrough.

Three levels of gas physics — Newton (microscopic), Boltzmann (mesoscopic), and Navier-Stokes (macroscopic) — bridged in a 2025 proof.

In 1900, the German mathematician David Hilbert set mathematics a challenge that has defined a century of research. He published a list of 23 unsolved problems — a kind of to-do list for the twentieth century and beyond. His sixth problem was among the boldest: can the laws of physics, which physicists derive from observation and intuition, be derived from pure mathematical axioms, the way theorems are proved from first principles?

One specific version of this challenge concerns the behaviour of gases. Physicists describe gas at three different scales: at the microscopic level, they use Newton’s laws of motion to track individual molecules; at the mesoscopic level (vast numbers of molecules but not yet a bulk fluid), they use the Boltzmann equation; and at the macroscopic level (a room full of air), they use the Navier-Stokes equations of fluid dynamics. All three describe the same physical reality, but they are not mathematically unified. Nobody had been able to rigorously derive one from another, from first principles, without patching in extra assumptions.

The 2025 breakthrough

In 2025, a trio of mathematicians — Yu Deng (University of Chicago), Zaher Hani (University of Michigan), and Xiao Ma — published a landmark proof connecting these three levels of description. Their work shows, rigorously and for the first time, how the microscopic Newtonian world of individual gas molecules transitions continuously into the Boltzmann equation and then into the macroscopic Navier-Stokes equations. No extra assumptions. No mathematical hand-waving.

“It reshapes our understanding of the natural world — not just solving a century-old problem, but providing a new mathematical foundation for the physics of fluids.” — Quanta Magazine, 2025

The proof is a tour de force of modern analysis, drawing on techniques from probability theory, kinetic theory, and partial differential equations. Quanta Magazine named it one of the top three mathematical breakthroughs of 2025, alongside the Kakeya conjecture and new results on hyperbolic surfaces.

Why it matters

The practical implications are significant. The Navier-Stokes equations underpin everything from weather forecasting and aircraft design to the modelling of ocean currents. Having a mathematically rigorous derivation of these equations from first principles does not change the equations themselves — but it gives physicists and mathematicians a far deeper understanding of when and why these equations can be trusted, and where their limits lie. It is the difference between knowing a recipe works and understanding the chemistry behind why it works.

Sources & Further Reading

Deng, Y., Hani, Z. & Ma, X. (2025). Full derivation of the Euler and Navier-Stokes equations from classical mechanics. Preprint.

Quanta Magazine (2025). The Biggest Breakthroughs in Mathematics: 2025. quantamagazine.org

GIGAZINE (2025). What are the three biggest breakthroughs in mathematics in 2025? gigazine.net

Scientific American (2025). The Top 10 Math Discoveries of 2025. scientificamerican.com

How a Korean mathematician solved the problem of moving furniture around a corner — a question that stumped the world since 1966.

The optimal “Gerver sofa” shape, with area 2.2195 square units — proven definitively by Jineon Baek in 2024 (arXiv:2411.19826)

Anyone who has ever wrestled a sofa around a tight corner knows the frustration. Mathematicians, it turns out, have been equally frustrated by the theoretical version of this problem since 1966, when Canadian mathematician Leo Moser posed it formally: what is the largest two-dimensional shape that can be manoeuvred around a right-angled corner in a hallway of unit width?

The question is deceptively simple. A plain square can make it around the corner, as can a semicircle. But what is the absolute maximum area any shape could have while still fitting through? For nearly six decades, mathematicians circled this problem (sometimes literally), proposing shapes and proving partial results, but never settling on a definitive answer.

Gerver’s candidate — and the missing proof

In 1992, mathematician Joseph Gerver proposed a beautiful, 18-curve shape that looked somewhat like the handset of a landline telephone. It had an area of approximately 2.2195 square units — an improvement on all previous candidates. But Gerver could not prove that nothing larger was possible. His shape sat as the best known answer for more than three decades: a champion with no certificate of victory.

“You keep holding on to hope, then breaking it, and moving forward by picking up ideas from the ashes.” — Jineon Baek

Baek’s 119-page proof

In late November 2024, Jineon Baek — a postdoctoral researcher at Yonsei University in Seoul — posted a 119-page paper to the preprint server arXiv that claimed to settle the matter. After seven years of work, Baek proved that no shape with an area larger than Gerver’s sofa can exist. The maximum sofa constant is 2.2195 square units, and that is final.

What is especially striking about Baek’s proof is that it relies entirely on logical reasoning — no large-scale computer simulations, no numerical approximations. Scientific American called it “surprising” that the final solution avoids computers altogether. The proof has been submitted to the prestigious Annals of Mathematics and is under peer review; early responses from leading geometers have been optimistic.

The moving sofa problem has a cultural footprint as well. The US sitcom Friends features a famous scene in which characters struggle to manoeuvre a sofa up a staircase while Ross Geller shouts “Pivot!” Scientific American noted, tongue in cheek, that “explaining the pivot required a 119-page paper.”

Sources & Further Reading

Baek, J. (2024). Optimality of Gerver’s Sofa. arXiv:2411.19826. arxiv.org

Scientific American (2025). Mathematicians Solve Infamous Moving Sofa Problem. scientificamerican.com

Phys.org (2024). Mathematician solves the moving sofa problem. phys.org

Korea Herald (2026). Six-decade math puzzle solved by Korean mathematician. koreaherald.com

Quanta Magazine (2025). The Largest Sofa You Can Move Around a Corner. quantamagazine.org

Google DeepMind’s AlphaProof reaches Olympiad level — and what it means for the future of mathematical proof.

AlphaProof’s performance at the 2024 International Mathematical Olympiad, Nature (2025)

Every summer, the most gifted young mathematicians in the world converge to compete in the International Mathematical Olympiad (IMO) — six problems, nine hours across two days, and a level of difficulty that humbles even exceptional minds. In 2024, an uninvited but remarkable competitor quietly entered the fray: Google DeepMind’s AlphaProof. It scored 28 out of a possible 42 points, placing it squarely in the silver-medal category and just one point short of gold.

What makes this achievement genuinely historic is not that a computer solved hard maths problems — machines have done arithmetic faster than humans for decades. What is new is that AlphaProof proved its answers with 100% verified correctness, step by logical step, using the formal proof assistant Lean. No guessing. No hallucinating a plausible-sounding answer. Every claim in every solution was machine-verified to be logically airtight.

How AlphaProof works

AlphaProof is a neuro-symbolic hybrid — it combines the intuitive pattern-matching of a large language model (based on Google’s Gemini architecture) with the rigorous verification engine of Lean, a formal mathematics software environment. The system was trained in three stages: first absorbing 300 billion tokens of mathematical text and code; then learning from 300,000 expert-written formal proofs; and finally, tackling 80 million formal mathematics problems through reinforcement learning — rewarded for every successful proof it constructed.

At the IMO, AlphaProof solved three non-geometry problems, including what IMO judges considered the hardest problem in the entire competition — a problem solved by only five human contestants. AlphaGeometry 2, a companion system, solved the geometry problem independently.

“AlphaProof is designed to prove mathematical statements — and it guarantees 100% correct solutions by verifying every logical step.” — Nature, 2025

Gold in 2025

The progression was swift. By 2025, an advanced version of Gemini equipped with “Deep Think” reasoning achieved the full gold-medal standard at IMO 2025 — solving all six problems within the time allowed. Unlike the 2024 effort, which required human experts to manually translate problems into formal language, the 2025 system could read and interpret problems in natural language directly. What previously demanded two to three days of computation was achieved within the competition’s standard timeframe.

This is not merely a parlour trick. Mathematicians across fields are beginning to ask: if AI can verify proofs and explore mathematical territories at machine speed, could it help crack problems that have resisted human effort for centuries — the Riemann Hypothesis, the Birch and Swinnerton-Dyer conjecture, P versus NP? AlphaProof does not answer these questions, but it brings them closer to the horizon than at any point in history.

Sources & Further Reading

Hubert, T. et al. (2025). Olympiad-level formal mathematical reasoning with reinforcement learning. Nature. DOI: 10.1038/s41586-025-09833-y

Google DeepMind (2025). AI achieves silver-medal standard solving International Mathematical Olympiad problems. deepmind.google

Google DeepMind (2026). Advanced version of Gemini with Deep Think officially achieves gold-medal standard at the IMO. deepmind.google

Phys.org (2025). AI math genius delivers 100% accurate results. phys.org/news/2025-11-ai-math-genius-accurate-results.html

Jenny Sturm/Shutterstock

Recently, a cheerful 100-year-old message in a bottle was found on the south-west coast of Australia. In it, a world war one soldier proclaimed to be “as happy as Larry”.

If you’re a betting person, you probably wouldn’t expect great odds of this happening. A bottle cast into the ocean could end up absolutely anywhere.

If it floats to a remote location, there is little chance of somebody stumbling upon it. And if it lands somewhere more favourable where people could potentially find it, there are other issues. The message itself will deteriorate over time as light degrades it. If the bottle fills with water, it will sink and almost certainly never be found.

So, what are the chances of a message in a bottle being found and it being over 100? And what are your chances of finding this bottle?

Despite these many possibilities during a bottle’s lifetime, the probability we are after is a straightforward calculation. Just count up the number of bottles with messages that have been found and are over 100 years old, and divide by the number of messages that have been sent this way (assuming we know how many are sent):

Probability calculation.

Our diagram below shows a hypothetical situation where 20 bottles are sent in total, of which six are found (indicated in gold) and one of these is over 100 years old (indicated by the “100” stamp). So, one in 20 bottles are found and over 100 years old. (Note: This is only a hypothetical calculation, not the real data.)

Hypothetical bottle data. Bottle image from https://www.flaticon.com/free-icons/bottle.

Instead of calculating the probability directly, another way to do it is by breaking the problem into two parts: (A) a bottle with a message is found, and (B) the found bottle is over 100. These two probabilities can be calculated separately and multiplied together to get what we want:

Multiplication rule of probability.

This is known as the “multiplication rule” of probability, and we confirm from our hypothetical numbers that (6/20)×(1/6) = 1/20, as before.

Both approaches to calculating this probability are simple. However, the direct calculation requires knowing the total number of bottles sent out, which is very difficult to know in the real world.

The multiplication rule has the advantage that it breaks the calculation into two parts. We can tackle each separately, then bring the two results together to get the probability we want. This is useful in the real-world situation where we can draw information from different sources.

First, we’ll deal with the probability that a bottle with a message is found, irrespective of its age.

Experts from the Federal Maritime and Hydrographic Agency of Germany suggest a one in ten chance that a message in a bottle will be found. This aligns broadly with various historical “drift bottle” experiments, where oceanographers released large numbers of bottles to understand ocean currents.

For example, studies from the 1960s and ’70s in the North Atlantic Ocean led to recovery rates of 14% from the Gulf of Mexico, 8% from the Caribbean Sea and 7% from the northern Brazilian coast. A more recent and more northerly study (between Canada and Greenland) from the 2000s led to a 5% recovery rate.

We would expect the results to vary naturally from different experiments in different parts of the world. But to keep things simple, we will stick with 1/10 as the probability that a bottle with a message is found.

Now for the second piece of the calculation: of the bottles that are found, what proportion are over 100 years old?

The table below summarises data from news articles collected on Wikipedia about very old bottles with messages that have been found. However, only data on bottles over 25 years old has been collected, presumably because older bottles are more newsworthy.

Data on the age distribution of bottles found, where the asterisk * indicates an estimated number.

So, we needed to estimate the number of 0- to 25-year-old bottles with messages ourselves – here’s how we did this.

The table shows that fewer bottles with messages are found as they get older. Messages in bottles degrade over time, which means the bottles have an increased chance of breaking and sinking, or just getting covered in layers of sediment. Plotting this data in the graph below helped us see the trend in the ages of found bottles more clearly.

Trend in the ages of bottles found.

We drew a line to match this observed trend in the ages of found bottles. This red line in the graph corresponds to the equation:

This equation provides an estimate of how many bottles have been found for any specific age range (where 25 = 0-to-25, 50 = 25-to-50 and so on). We are interested in the the 0- to 25-year-old bottles, so the equation suggests 46 bottles have been found in this range.

Adding up this and all of the numbers in the table gives a total of 106 bottles found, of which 12 are over 100 years old, and 12/106 is about one in ten.

Recapping the above, we have that: (A) one in ten bottles with messages are found, of which (B) one in ten are over 100 years old. Bringing these results together using the multiplication rule, we estimate the chance of a message in a bottle being found and it being over 100 years old to be (1/10)×(1/10) = 1/100.

So, if there are 100,000 bottles with messages floating around the oceans waiting to be found, we’d expect 1,000 of these to be found and be 100 or more years old. Assuming anybody in the world is equally likely to find one of these, with 8 billion people currently, that’s about a one in 8 million chance of you finding one – pretty unlikely.

However, some people are more persistent at message-in-a-bottle hunting than others. Following the paths of ocean currents (known as gyres) could provide clues on where to look.

Specifically, peninsulas or islands intersecting with these gyres could be good spots. For this reason, it has been suggested the Caribbean islands are ideally placed for finding bottles as they lie on the path of the North Atlantic Gyre. Which seems like a great reason to travel to the Carribean!

But let’s also spare a thought for the poor soul stranded on their desert island, who surely won’t appreciate the low odds of their SOS being found.

For more such insights, log into www.international-maths-challenge.com.

*Credit for article given to Kevin Burke & David O’Sullivan*

Typical algorithms for doing square roots by hand require estimation. I have taught a different algorithm that does not rely on estimation but instead uses subtraction of successive odd integers. First, I offer examples that illustrate two situations that may arise. Then I present a third situation (as well as how to deal with the square roots of non-perfect squares).

This approach is based on the fact that the nth perfect square is the sum of the first n odd integers. This fact can be used to subtract successive odd integers from a given number for which one wishes to find the square root. If the number isn’t a perfect square, this method can be extended by adding pairs of zeroes to the original number and continuing the process for each additional decimal place one wishes find.

FIRST RULE

It helps to look at a couple of examples to illustrate two “special cases” that arise with some numbers, requiring one or two additional “rules” or steps.

Using the example of 54,756:

Start by marking pairs of digits from the right-most digit: 5 | 47 | 56

Then subtract 1 from the leftmost digit or pair: 5 – 1 = 4.
Continue with the next odd integer: 4 – 3 = 1.

We can’t subtract 5 from 1, so we count how many odd integers we’ve subtracted thus far (2) and mark that above the 5.

Bring down the next pair of digits and append it to the 1 yielding 147.

To get the next odd integer to subtract, multiply the last odd integer subtracted by 10 and add 11 (this is Rule #1) to the product. Here, we have 10 x 3 + 11 = 41. Proceed as previously, subtracting 41 from 147 = 106.
Subtract the next odd integer, 43 from 106 = 63.
Subtract the next odd integer, 45 from 63 = 18.

Again, we can’t subtract 47 from 18; counting, we have done 3 subtractions and place 3 above the pair 47. Multiply 45 x 10 and add 11 = 461.

Bring down the next pair of digits, 56, and append them to the 18, yielding 1856.

Subtract 461 from 1856 = 1395.
Subtract the next odd integer, 463 from 1395 = 932.
Subtract the next odd integer, 465 from 932 = 467.
Subtract the next odd integer, 467 from 467 = 0.
Stop.

Counting how many subtractions, we see it is 4 and we write 4 above the 56.

Our answer is that 234 is the square root of 54,756. Alternately, instead of keeping a running total of the subtractions and placing the digits above successive pairs of digits from the left, take the last number subtracted, 467, add 1, and divide the result by 2 = 234, same as what we determined the other way.

SECOND RULE

A second example introduces another rule not previously required: find the square root of 4,121,062,016 using the subtraction of successive odd integers.

Begin as above by making pairs of digits from the right-most digit: 4 | 12 | 10 | 62 | 40 | 16

Subtract 1 from 4 = 3.

Subtract 3 from 3 = 0.
Write down 2 for the two subtractions above the 4.
Bring down the next pair of digits, 12.
Multiply 3 x 10 and add 11 = 41.
Note that 41 is too big to subtract from 12.
Write 0 above the 12, since we did 0 subtractions.

Bring down the next pair of digits, 10, and append to the 12 => 1210.

Insert a 0 to the left of the last digit in 41 => 401. (This is Rule #2)
Subtract 401 from 1210 = 809.
Subtract the next odd integer, 403, from 809 = 406.
Subtract the next odd integer, 405 from 406 = 1.

For the three subtractions, write 3 above the 10.

Bring down the next pair of integers, 62 and append to the 1 => 162
Multiply 405 by 10 and add 11 = 4061.
We need to apply Rule #2 again. Write 0 above the 62, bring down the next pair of digits, 40, and append to the 162 => 16240.
Insert 0 to the left of the last digit of 4061 => 40601.

Note that this is still too big to subtract from 16240.
Apply Rule #2 again (and it may have to be applied more than twice in particular cases).
Write 0 above the 40, bring down the 16 and append to the 16240 => 1624016.
Insert a 0 to the left of the last digit of 40601 => 406001.

Subtract 406001 from 1624016 = 1218015.
Subtract the next odd integer , 406003 from 1218015 = 812012.
Subtract the next odd integer, 406005 from 812012 = 406007.
Subtract the next odd integer, 406007 from 406007 = 0.
Write a 4 above the last pair of digits, 16.

The square root of 41210624016 = 203,004.

Again, alternately, the answer = (406007+1) / 2 = 203,004.

THIRD RULE

There is a group of numbers for which the process previously described won’t work. For example, try to use it to find the square root of 100.

Grouping as before: 1 | 00

Subtracting 1 from 1 = 0.

Write 1 above the 1, bring down the next pair of digits, 00, and append to the 0.

Multiply 1 x 10 and add 11 = 21.

Can’t subtract 21 from 0. Hmm. Although we know the answer is 10, to make things work, we can note the following, which is Rule #3:

If you want the square root of a whole number that ends in two or more zeros, write the number as a product of a number and an even power of ten.

So 100 = 1 x 10^2.

We get that the square root of 1 = 1, append one zero for every pair of zeroes in the original number, and Bob’s your uncle. (Or something like that).

For example, to find the square root of 3,610,000, remove two pairs of zeroes from the original number, then apply the original procedure:

Group: 3 | 61.

Subtract 1 from 3 = 2

Can’t subtract 3 from 2, so write 1 above the 3, bring down the next pair of digits and append them to the 2 => 261.

Multiply 1 x 10 and add 11 = 21.

Subtract 21 from 261 = 240.
Subtract 23 from 240 = 217
Subtract 25 from 217 = 192
Subtract 27 from 192 = 165
Subtract 29 from 165 = 136
Subtract 31 from 136 = 105
Subtract 33 from 105 = 72
Subtract 35 from 72 = 37
Subtract 37 from 37 = 0

So write a 9 above the 61. Append two zeroes to the 19, one for each pair removed.
Then the square root of 3,610,000 = 1900.

DEALING WITH NON-PERFECT SQUARES

Finally, this process works for whole numbers that aren’t perfect squares and for decimals. It just won’t terminate in those cases (except arbitrarily). For a decimal, also break the number into pairs of digits to the right of the decimal point.

For example, finding the square root of 3 to 3 decimal places.

Append pairs of zeroes for each decimal place you want in the answer, plus two more to be able to round to the given place.

So write 3 as 3 | 00 | 00 | 00 | 00

Subtract 1 from 3 = 2.

Write 1 above the 3. Bring down a pair of zeroes, append to the 2 => 200.

Multiply 1 x 10 and add 11 = 21.

Subtract 21 from 200 = 179
Subtract 23 from 179 = 156
Subtract 25 from 156 = 131
Subtract 27 from 131 = 104
Subtract 29 from 104 = 75
Subtract 31 from 75 = 44
Subtract 33 from 44 = 11.

Write 7 above the first pair of zeroes.

Bring down the next pair of zeroes and append to the 11 => 1100.

Multiply 33 x 10 and add 11 = 341.

Subtract 341 from 1100 = 759.
Subtract 343 from 759 = 416.
Subtract 345 from 416 = 71.

Write 3 above the second pair of zeroes.

Append the next pair of zeroes to the 71 => 7100.

Multiply 345 x 10 and add 11 = 3461.

Subtract 3461 from 7100 = 3639.
Subtract 3463 from 3639 = 176.

Write 2 above the third pair of zeroes.

Append the last pair of zeroes to the 176 => 17600

Multiply 3463 x 10 and add 11 = 346241.

We could continue, but it suffices to realize that the next digit will be 0 and so our answer is that the square root of 3 is 1.732 rounded to three decimal places.

For more such insights, log into www.international-maths-challenge.com.

*Credit for article given to Michael Goldenberg*

I’d never heard of this thing until grad school. And even then, I never asked what it was. Over the course of time I eventually figured it out, but never really got an opportunity to do much with it. Nor have I had a chance to teach it.

A teacher interview question from Oleg Gleizer’s book inspired me to think about, and learn, this nifty skill.

So what is it?

Here’s the definition (mostly from Wikipedia):

A ruler-and-compass construction is the construction of lengths, angles, and geometric figures using only a ruler and compass.

This means that you can take one of those “pointer and pencil circle making things” and anything really straight (the side of your new iPhone, the edge of a file folder, etc.) and make pretty much create anything in geometry.

Pretty cool, huh?

I gave it a shot!

I used Oleg’s teacher interview question:

Given a straight line and a point away from it, how would you draw another straight line passing through the point and perpendicular to the original line, using a compass and straightedge as tools?

Can I do it? Of course!

Well… I thought about it and it seemed like I could. So I went out and got a compass, and used a fingernail file as a straight edge. Here’s how I did it:

Here’s the line and the point. Easy peasy.

I made an arc from the point through the line, so I would have two spots on the line (where the circle piece went through):

From those two places, I made two more arcs through the point above and long enough to run into each other below:

I connected the point with the intersection of the arcs at the bottom and VOILA: perpendicular line to the other line!

Join me in the journey!

This is the first in my ruler and compass journey. They’re kind of fun, and I want to do more. So I will house them here, for future reference.

Here are the first 10 on my list.

1. Line perpendicular to given line through given point not on given line. (this one)
2. Perpendicular bisector of given segment.
3. Right angle at given point on given line.
4. Square with given segment as side.
5. Equilateral triangle with given segment as side.
6. Hexagon with given segment as side.
7. Copy a given angle to a given segment.
8. Line parallel to given line through point not on given line.
9. Dividing given segment into N equal parts.
10. Bisecting a given angle.

Grab a straightedge and compass for each member of your family – let me know you’re on board in the comments.

For more such insights, log into www.international-maths-challenge.com.

*Credit for article given to Bon Crowder*

Dan Smith

Warning this article contains spoilers about the new Amazon Prime series Young Sherlock.

I’ve read the whole Sherlock Holmes canon multiple times over. I love how Holmes uses analytical reasoning to unravel problems that look mysterious, but ultimately prove to have simple explanations. So I was excited when I saw Guy Ritchie’s Young Sherlock appear on Amazon Prime. My excitement was quickly tempered when I started watching, though.

A key part of the plot relies on mathematics. Holmes first meets his sidekick Moriarty (yes, he is working together with his future adversary) at the blackboard after a maths lecture at Oxford. Despite some mistakes in the dialogue, the maths on the blackboard is interesting enough. It is finding the solutions to the equation x5 + x4 + x3 + x2 + x + 1 = 0.

In the maths many of us will have learned at school, we are taught that a positive times a positive makes a positive and that a negative times a negative also makes a positive. For example, 3 times 3 equals 9, but -3 times -3 also equals 9. Squaring a number (when you multiply a number by itself) should always give a positive result. The reverse operation – finding the number(s) you multiply together to give a positive number – is called taking the square root. The two square roots of 9 are 3 and -3, since when you square either of these numbers you get the answer 9.

If we want to take the square root of -1, say, then we need to venture into the realm of imaginary numbers. Imaginary numbers are the square roots of negative numbers. Mathematicians defined the imaginary number i to be the square root of -1 (technically -1 has two square roots i and -i). The square roots of other negative numbers are multiples of i. The square roots of -9, for example are 3i and -3i. Some of the solutions from the equation on the blackboard involve imaginary numbers (this will turn out to be an important plot point).

Mathematical blunders

It’s plausible that the equation on the blackboard might appear in an early first year undergraduate tutorial. Something approaching a passable solution is given, but in excruciating detail (the sort of detail you wouldn’t use at school, let alone in a maths degree at Oxford). And there are mistakes in the maths.

Young Sherlock Holmes contemplates the incorrect solutions on the blackboard. Amazon Prime screenshot

Towards the end of the lecture, the professor sets the students homework to find all the solutions to the equation, even though they are already written on the board (although incorrectly). Despite this, the end of the scene sees Sherlock spending some time trying to think of the solutions before Moriarty comes up and shows him two of the five solutions (as if they were the only ones). Moriarty too writes these down incorrectly, but in a different way to the incorrectness already on the board.

As Moriarty writes down the complex solution (complex means the answer contains both real and imaginary numbers) he says “These solutions, they’re not real. They’re imaginary.” which we can allow (although technically he means complex).

What we can’t forgive is Moriarty going on to say, “That means even if you can’t see the target, you can still shoot for it.” Which is nonsense, even as a metaphor. Complex numbers aren’t targets you can’t see, but well-defined, mainstream (even in the 1870s) mathematical quantities and there’s no sense in which you “aim at” a complex solution to an equation.

Death by numbers

In the last episode, Holmes and his team are battling to halt the distribution of a deadly chemical weapon known as the “creeping death”. They find a scrap of paper in a secret room which they say is the “equation for creating the creeping death.”

I was expecting to see some complex chemical reaction formulae sketched on the page, but when it’s held up to the camera, we see instead a mathematical equation: z3 + 4 z2 – 10 z + 12 = 0.

What does this have to do with the chemical process for creating the deadly nerve agent?

Nothing, it turns out. Or at least nothing I can imagine. In fact it’s a device to allow Holmes and Moriarty to hark back to that moment in the lecture theatre when they first met. What follows goes beyond artistic license into the realm of gibberish.

“If we have the positive equation”, they say, “then we can come up with the negative. And thus create a compound to neutralise the threat of creeping death.” Perhaps they meant “positive solution”, because equations themselves aren’t positive or negative. Either way, the idea that this simple mathematical equation or its solutions are the secret formula for making a weapon of mass destruction doesn’t make sense. There’s no context, no sense in which this equation could be the secret recipe for creating the nerve agent.

Moriarty points out that they have a problem. “This equation is not finished.” By this I think he means that the three solutions to the equation are not written out explicitly.

One solution, z = – 6 is given. And it’s correct. The rest of the scrap of paper contains a reformulation of the equation (a factorisation), which shows that the remaining solutions can be found by solving a quadratic equation: z2 – 2 z + 2 = 0.

A quadratic equation is just an equation built around a squared term (in this case z2 ), which has two solutions. The formula for the solutions may be familiar to GCSE students (normally aged 15 to 17 years old). For a general quadratic equation: a z2 + b z + c = 0, the two solutions are given below.

Yet, we are supposed to believe that, despite having supposedly solved a far more complicated equation than this in the first episode, Moriarty can’t find the solution to this much simpler equation. So stumped is Moriarty – the future maths professor – that he spends precious time, as a bomb is about to detonate, searching for a piece of paper with this missing solution. He almost loses his life when he could have just used a GCSE-level formula.

The piece of paper he eventually finds contains an incorrect statement of the quadratic formula alongside some nonsensical text, although the solutions are at least correct: z = 1 + i and z = 1 – i (where i, remember, is the imaginary number).

I appreciate my dissection of the maths is high-grade nerdery. Most people will have watched the series without pausing it like I did to look at the maths and probably won’t have noticed. But, if maths is going to be a pivotal plot point in your blockbuster series, then you’ll probably want to make sure you get it right.

For more such insights, log into www.international-maths-challenge.com.

*Credit for article given to Kit Yates*

Keeping children engaged with maths over the summer doesn’t have to be complicated. Grow their maths skills alongside your very own garden with these tips.

As some children continue to learn from home, maintaining their mathematical skills can be a challenge. Parents and caregivers may feel nervous about teaching maths to their children, and even hold onto some maths anxiety themselves.

It’s okay to take a simple approach to maths teaching using objects and environments at home. Have you considered that gardening can be an easy and effective way to help your child connect with mathematical concepts in a concrete and lasting way?

Here are four ways you can support teaching maths in the garden.

1. Invite your child to help with garden planning

Maths is everywhere! And your garden is no exception. Creating gardening maths activities can be as simple as taking notice of what you’re already doing.

Ready to plan your garden? Start by breaking down the information you need. The planning stage of gardening is full of rich mathematical calculations like:

How much space do we have to plant?

How much soil will we need?

How much will it cost?

How far apart will we need to plant the different varieties of flowers or vegetables?

How many rows do we have space to plant? How many columns?

How much sunlight does this spot get? How much sun will our plants need?

To an expert gardener, these may seem like simple things you think about in passing. But they can easily become mathematical tasks you could get your child to help you with.

2. Ask your child to help you find the answers

You need information to get your garden going, and you have a helper ready to get it for you. Ask your child questions that will help you prepare for planting:

Can they measure the length and width of your garden?

Can they find the area?

If it’s a raised garden bed, can they find the height and the volume?

If they know the volume in cubic metres or centimetres, can they express it in litres?

If a standard bag of soil is 50 l and costs £12, can they tell you how many bags you’ll need to fill the garden bed?

Can they tell you how much the soil will cost overall?

Posing mathematical questions that are rooted in reality gives your child and opportunity to use their knowledge. When your child can see why they’re doing something, they develop a deeper conceptual understanding.

3. Challenge your child to find multiple solutions

If you think your child could go a bit further, ask them to plot out where to plant different varieties in your garden. Say you want to plant lettuce, beetroot and carrots — and they all need to be planted the following distances away from other plants:

Lettuce should be planted at least 30 cm away from other plants

Beetroot should be planted at least 10 cm away from other plants

Carrots should be planted at least 5 cm away from other plants

With available garden space in mind, you can ask guiding questions like:

How many configurations could you have?

Could you plant 5 lettuce, 10 beetroot and 20 carrots?

Or could you plant more lettuce and fewer beetroot and carrots?

 

Ask your child to map it out, and come up with a few different answers. Finding more than one way to solve a problem will boost your child’s reasoning skills, allow them to explore maths for themselves and encourage creativity!

4. Continue learning by encouraging maths journaling

Keeping a garden is a great opportunity for your child to reflect on what they learned in an ongoing maths journal. For example, when your child thinks back on what they did to measure and plan the garden, you could ask:

Why did we need that information?

What did it help us do next?

When planting, they can keep a record of what, where, when and how many seeds were planted. Have them make estimations like:

How many plants do you think will grow?

What do you think the yield will be?

When harvesting, encourage them to refer back to their planting journal and compare.

Can they compare their estimations to the actual yield?

Can they compare the actual yield to the data they recorded when planting?

Can they tell you what percentage of seeds grew into plants?

How could they use their findings in the future?

Reflecting on their learning and answering open questions will help your child master mathematical concepts in depth. Explaining their thinking, getting creative and making connections to what they learned and why, all help to solidify mathematical understanding.

In short, learning maths at home doesn’t need to be complicated. Teaching mathematics can also be a part of teaching your child life lessons and skills, like how to plant a garden.

Learning maths in real-life contexts helps children form a connection with new information, and better understand how to apply it. So they won’t just understand what to do, they’ll understand why they’re doing it.

For more such insights, log into www.international-maths-challenge.com.

*Credit for article given to Lisa Champagne*